f(x)=ln(x). Find the area underneath the curve f(x) between 1 and 2.
We cannot dirrectly intergrate ln(x), so instead we intergrate 1ln(x) using intergration by parts.
The formula for intergration by parts is: ∫ (udv/dx) dx = uv − ∫ vdu/dx dx .
We let u=ln(x) so that du/dx=1/xWe let dv/dx=1 so that v=x
We put those values into the formula and we get ∫ ln(x) dx = xln(x) - ∫ (x1/x )dx∫ ln(x) dx = xln(x) - ∫1 dx∫ ln(x) dx = xln(x)-x + c
Finding the area under the curve between 1 and 2. ∫21 ln(x) dx = [xln(x)-x]21 ∫21 ln(x) dx = 2ln(2)-2-(1ln(1)-1) ∫21 ln(x) dx = 2*ln(2)-1
AT
Answered by Angus T. • Further Mathematics tutor
3488 Views
See similar Further Mathematics A Level tutorsRelated Further Mathematics A Level answers
All answers ▸You are given a polynomial f, where f(x)=x^4 - 14x^3 + 74 x^2 -184x + 208, you are told that f(5+i)=0. Express f as the product of two quadratic polynomials and state all roots of f.
Let I(n) = integral from 1 to e of (ln(x)^n)/(x^2) dx where n is a natural number. Firstly find I(0). Show that I(n) = -(1/e) + n*I(n-1). Using this formula find I(1).
Why does e^ix = cos(x) + isin(x)
Find the root of the complex 3+4i